Request:
Please check my work. State where errors, if any, occurred and how to correct them. Is there a better way to calculate the transform other than the present method given?
Given:
Find the Laplace transform $\mathcal{L}\{t\cdot e^{-t}\cdot \cos(t)\}$ by carefully selecting and differentiating $f(t)$.
Solution:
$$\mathcal{L}\{t\cdot f(t)\}=-F'(s)$$
$$f(t)=e^{-t}\cdot \cos(t)$$
$$F(s)=\frac{1+s}{(s+1)^2+1}$$
$$F'(s)=\frac{(-1)\cdot(2s+2)(1+s)}{[(s+1)^2+1]^2}$$
$$F'(s)=-\frac{2(s^2+2s+1)}{[(s+1)^2+1]^2}$$
Answer:
$$\mathcal{L}\{t\cdot e^{-t}\cdot \cos(t)\}=\frac{2(s^2+2s+1)}{[(s+1)^2+1]^2}$$